3.78.7 \(\int \frac {-4+(4 x-4 e x-6 x^2) \log (x^2)+(-2 x+2 e x+6 x^2) \log ^2(x^2)+(2 x^2+2 e^2 x^2-9 x^3+9 x^4+e (-4 x^2+9 x^3)) \log ^3(x^2)}{x \log ^3(x^2)} \, dx\)

Optimal. Leaf size=24 \[ x^2 \left (-1+e+\frac {3 x}{2}+\frac {1}{x \log \left (x^2\right )}\right )^2 \]

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Rubi [B]  time = 0.30, antiderivative size = 55, normalized size of antiderivative = 2.29, number of steps used = 21, number of rules used = 10, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {6688, 14, 2302, 30, 2320, 2330, 2300, 2178, 2307, 2298} \begin {gather*} \frac {9 x^4}{4}-3 (1-e) x^3+(1-e)^2 x^2+\frac {1}{\log ^2\left (x^2\right )}-\frac {(2 (1-e)-3 x) x}{\log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + (4*x - 4*E*x - 6*x^2)*Log[x^2] + (-2*x + 2*E*x + 6*x^2)*Log[x^2]^2 + (2*x^2 + 2*E^2*x^2 - 9*x^3 + 9*
x^4 + E*(-4*x^2 + 9*x^3))*Log[x^2]^3)/(x*Log[x^2]^3),x]

[Out]

(1 - E)^2*x^2 - 3*(1 - E)*x^3 + (9*x^4)/4 + Log[x^2]^(-2) - ((2*(1 - E) - 3*x)*x)/Log[x^2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2300

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2307

Int[(x_)^(m_.)/Log[(c_.)*(x_)^(n_)], x_Symbol] :> Dist[1/n, Subst[Int[1/Log[c*x], x], x, x^n], x] /; FreeQ[{c,
m, n}, x] && EqQ[m, n - 1]

Rule 2320

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(x*(d + e*x)^q*(a
+ b*Log[c*x^n])^(p + 1))/(b*n*(p + 1)), x] + (-Dist[(q + 1)/(b*n*(p + 1)), Int[(d + e*x)^q*(a + b*Log[c*x^n])^
(p + 1), x], x] + Dist[(d*q)/(b*n*(p + 1)), Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^(p + 1), x], x]) /; FreeQ
[{a, b, c, d, e, n}, x] && LtQ[p, -1] && GtQ[q, 0]

Rule 2330

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (x \left (2 (1-e)^2-9 (1-e) x+9 x^2\right )-\frac {4}{x \log ^3\left (x^2\right )}+\frac {4-4 e-6 x}{\log ^2\left (x^2\right )}+\frac {2 (-1+e+3 x)}{\log \left (x^2\right )}\right ) \, dx\\ &=2 \int \frac {-1+e+3 x}{\log \left (x^2\right )} \, dx-4 \int \frac {1}{x \log ^3\left (x^2\right )} \, dx+\int x \left (2 (1-e)^2-9 (1-e) x+9 x^2\right ) \, dx+\int \frac {4-4 e-6 x}{\log ^2\left (x^2\right )} \, dx\\ &=-\frac {(2 (1-e)-3 x) x}{\log \left (x^2\right )}+2 \int \left (\frac {-1+e}{\log \left (x^2\right )}+\frac {3 x}{\log \left (x^2\right )}\right ) \, dx-2 \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log \left (x^2\right )\right )-(2 (1-e)) \int \frac {1}{\log \left (x^2\right )} \, dx+\int \left (2 (-1+e)^2 x+9 (-1+e) x^2+9 x^3\right ) \, dx+\int \frac {4-4 e-6 x}{\log \left (x^2\right )} \, dx\\ &=(1-e)^2 x^2-3 (1-e) x^3+\frac {9 x^4}{4}+\frac {1}{\log ^2\left (x^2\right )}-\frac {(2 (1-e)-3 x) x}{\log \left (x^2\right )}+6 \int \frac {x}{\log \left (x^2\right )} \, dx-(2 (1-e)) \int \frac {1}{\log \left (x^2\right )} \, dx-\frac {((1-e) x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{\sqrt {x^2}}+\int \left (\frac {4 (1-e)}{\log \left (x^2\right )}-\frac {6 x}{\log \left (x^2\right )}\right ) \, dx\\ &=(1-e)^2 x^2-3 (1-e) x^3+\frac {9 x^4}{4}-\frac {(1-e) x \text {Ei}\left (\frac {\log \left (x^2\right )}{2}\right )}{\sqrt {x^2}}+\frac {1}{\log ^2\left (x^2\right )}-\frac {(2 (1-e)-3 x) x}{\log \left (x^2\right )}+3 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,x^2\right )-6 \int \frac {x}{\log \left (x^2\right )} \, dx+(4 (1-e)) \int \frac {1}{\log \left (x^2\right )} \, dx-\frac {((1-e) x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{\sqrt {x^2}}\\ &=(1-e)^2 x^2-3 (1-e) x^3+\frac {9 x^4}{4}-\frac {2 (1-e) x \text {Ei}\left (\frac {\log \left (x^2\right )}{2}\right )}{\sqrt {x^2}}+\frac {1}{\log ^2\left (x^2\right )}-\frac {(2 (1-e)-3 x) x}{\log \left (x^2\right )}+3 \text {li}\left (x^2\right )-3 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,x^2\right )+\frac {(2 (1-e) x) \operatorname {Subst}\left (\int \frac {e^{x/2}}{x} \, dx,x,\log \left (x^2\right )\right )}{\sqrt {x^2}}\\ &=(1-e)^2 x^2-3 (1-e) x^3+\frac {9 x^4}{4}+\frac {1}{\log ^2\left (x^2\right )}-\frac {(2 (1-e)-3 x) x}{\log \left (x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 28, normalized size = 1.17 \begin {gather*} \frac {\left (2+x (-2+2 e+3 x) \log \left (x^2\right )\right )^2}{4 \log ^2\left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + (4*x - 4*E*x - 6*x^2)*Log[x^2] + (-2*x + 2*E*x + 6*x^2)*Log[x^2]^2 + (2*x^2 + 2*E^2*x^2 - 9*x^
3 + 9*x^4 + E*(-4*x^2 + 9*x^3))*Log[x^2]^3)/(x*Log[x^2]^3),x]

[Out]

(2 + x*(-2 + 2*E + 3*x)*Log[x^2])^2/(4*Log[x^2]^2)

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fricas [B]  time = 0.64, size = 75, normalized size = 3.12 \begin {gather*} \frac {{\left (9 \, x^{4} - 12 \, x^{3} + 4 \, x^{2} e^{2} + 4 \, x^{2} + 4 \, {\left (3 \, x^{3} - 2 \, x^{2}\right )} e\right )} \log \left (x^{2}\right )^{2} + 4 \, {\left (3 \, x^{2} + 2 \, x e - 2 \, x\right )} \log \left (x^{2}\right ) + 4}{4 \, \log \left (x^{2}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2*exp(1)^2+(9*x^3-4*x^2)*exp(1)+9*x^4-9*x^3+2*x^2)*log(x^2)^3+(2*x*exp(1)+6*x^2-2*x)*log(x^2)^
2+(-4*x*exp(1)-6*x^2+4*x)*log(x^2)-4)/x/log(x^2)^3,x, algorithm="fricas")

[Out]

1/4*((9*x^4 - 12*x^3 + 4*x^2*e^2 + 4*x^2 + 4*(3*x^3 - 2*x^2)*e)*log(x^2)^2 + 4*(3*x^2 + 2*x*e - 2*x)*log(x^2)
+ 4)/log(x^2)^2

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giac [B]  time = 0.20, size = 107, normalized size = 4.46 \begin {gather*} \frac {9 \, x^{4} \log \left (x^{2}\right )^{2} + 12 \, x^{3} e \log \left (x^{2}\right )^{2} - 12 \, x^{3} \log \left (x^{2}\right )^{2} + 4 \, x^{2} e^{2} \log \left (x^{2}\right )^{2} - 8 \, x^{2} e \log \left (x^{2}\right )^{2} + 4 \, x^{2} \log \left (x^{2}\right )^{2} + 12 \, x^{2} \log \left (x^{2}\right ) + 8 \, x e \log \left (x^{2}\right ) - 8 \, x \log \left (x^{2}\right ) + 4}{4 \, \log \left (x^{2}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2*exp(1)^2+(9*x^3-4*x^2)*exp(1)+9*x^4-9*x^3+2*x^2)*log(x^2)^3+(2*x*exp(1)+6*x^2-2*x)*log(x^2)^
2+(-4*x*exp(1)-6*x^2+4*x)*log(x^2)-4)/x/log(x^2)^3,x, algorithm="giac")

[Out]

1/4*(9*x^4*log(x^2)^2 + 12*x^3*e*log(x^2)^2 - 12*x^3*log(x^2)^2 + 4*x^2*e^2*log(x^2)^2 - 8*x^2*e*log(x^2)^2 +
4*x^2*log(x^2)^2 + 12*x^2*log(x^2) + 8*x*e*log(x^2) - 8*x*log(x^2) + 4)/log(x^2)^2

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maple [B]  time = 0.06, size = 69, normalized size = 2.88

method	result	size
risch	\(\frac {9 x^{4}}{4}+3 x^{3} {\mathrm e}-3 x^{3}+x^{2} {\mathrm e}^{2}-2 x^{2} {\mathrm e}+x^{2}+\frac {2 x \,{\mathrm e} \ln \left (x^{2}\right )+3 x^{2} \ln \left (x^{2}\right )-2 x \ln \left (x^{2}\right )+1}{\ln \left (x^{2}\right )^{2}}\)	\(69\)
norman	\(\frac {1+\left (-3+3 \,{\mathrm e}\right ) x^{3} \ln \left (x^{2}\right )^{2}+\left (2 \,{\mathrm e}-2\right ) x \ln \left (x^{2}\right )+\left ({\mathrm e}^{2}-2 \,{\mathrm e}+1\right ) x^{2} \ln \left (x^{2}\right )^{2}+3 x^{2} \ln \left (x^{2}\right )+\frac {9 x^{4} \ln \left (x^{2}\right )^{2}}{4}}{\ln \left (x^{2}\right )^{2}}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2*exp(1)^2+(9*x^3-4*x^2)*exp(1)+9*x^4-9*x^3+2*x^2)*ln(x^2)^3+(2*x*exp(1)+6*x^2-2*x)*ln(x^2)^2+(-4*x*
exp(1)-6*x^2+4*x)*ln(x^2)-4)/x/ln(x^2)^3,x,method=_RETURNVERBOSE)

[Out]

9/4*x^4+3*x^3*exp(1)-3*x^3+x^2*exp(2)-2*x^2*exp(1)+x^2+(2*x*exp(1)*ln(x^2)+3*x^2*ln(x^2)-2*x*ln(x^2)+1)/ln(x^2
)^2

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maxima [B]  time = 0.40, size = 59, normalized size = 2.46 \begin {gather*} \frac {9}{4} \, x^{4} + 3 \, x^{3} e - 3 \, x^{3} + x^{2} e^{2} - 2 \, x^{2} e + x^{2} + \frac {3 \, x^{2} + 2 \, x {\left (e - 1\right )}}{2 \, \log \relax (x)} + \frac {1}{\log \left (x^{2}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2*exp(1)^2+(9*x^3-4*x^2)*exp(1)+9*x^4-9*x^3+2*x^2)*log(x^2)^3+(2*x*exp(1)+6*x^2-2*x)*log(x^2)^
2+(-4*x*exp(1)-6*x^2+4*x)*log(x^2)-4)/x/log(x^2)^3,x, algorithm="maxima")

[Out]

9/4*x^4 + 3*x^3*e - 3*x^3 + x^2*e^2 - 2*x^2*e + x^2 + 1/2*(3*x^2 + 2*x*(e - 1))/log(x) + 1/log(x^2)^2

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mupad [B]  time = 6.29, size = 47, normalized size = 1.96 \begin {gather*} \frac {{\left (2\,x\,\mathrm {e}-2\,x+3\,x^2\right )}^2}{4}+\frac {\ln \left (x^2\right )\,\left (2\,x\,\mathrm {e}-2\,x+3\,x^2\right )+1}{{\ln \left (x^2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2)^3*(2*x^2*exp(2) - exp(1)*(4*x^2 - 9*x^3) + 2*x^2 - 9*x^3 + 9*x^4) - log(x^2)*(4*x*exp(1) - 4*x +
6*x^2) + log(x^2)^2*(2*x*exp(1) - 2*x + 6*x^2) - 4)/(x*log(x^2)^3),x)

[Out]

(2*x*exp(1) - 2*x + 3*x^2)^2/4 + (log(x^2)*(2*x*exp(1) - 2*x + 3*x^2) + 1)/log(x^2)^2

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sympy [B]  time = 0.15, size = 58, normalized size = 2.42 \begin {gather*} \frac {9 x^{4}}{4} + x^{3} \left (-3 + 3 e\right ) + x^{2} \left (- 2 e + 1 + e^{2}\right ) + \frac {\left (3 x^{2} - 2 x + 2 e x\right ) \log {\left (x^{2} \right )} + 1}{\log {\left (x^{2} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2*exp(1)**2+(9*x**3-4*x**2)*exp(1)+9*x**4-9*x**3+2*x**2)*ln(x**2)**3+(2*x*exp(1)+6*x**2-2*x)*
ln(x**2)**2+(-4*x*exp(1)-6*x**2+4*x)*ln(x**2)-4)/x/ln(x**2)**3,x)

[Out]

9*x**4/4 + x**3*(-3 + 3*E) + x**2*(-2*E + 1 + exp(2)) + ((3*x**2 - 2*x + 2*E*x)*log(x**2) + 1)/log(x**2)**2

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